A231693 -id:A231693 - cp's OEIS frontend

A231692 Define a sequence of rationals by f(0)=0, thereafter f(n)=f(n-1)-1/n if that is >= 0, otherwise f(n)=f(n-1)+1/n; a(n) = numerator of f(n).

0, 1, 1, 1, 5, 13, 1, 27, 19, 451, 199, 4709, 2399, 3467, 29207, 5183, 55411, 221267, 300649, 1628251, 5508127, 259001, 762881, 6460903, 5694791, 11476403, 27820203, 326206681, 5151783667, 69088293143, 146724611903, 2219373406193, 8951357840311, 4575492601111, 328329280711, 4454145077671

Offset: 0

N. J. A. Sloane, Nov 15 2013, Nov 16 2013

It is conjectured that the terms of the {f(n)} sequence are distinct.
If that is true, then the {f(n)} sequence is a fractional analog of Recamán's sequence A005132.
The denominators of {f(n)} form A231693 (a non-monotonic sequence).
From Don Reble, Nov 16 2013: (Start)
Here is a proof that the f(n) are distinct: Suppose not. Then the difference between the terms (which is zero) is a number of the form +- 1/(a+1) +- 1/(a+2) +- 1/(a+3) +- ... +- 1/(a+n).
Consider any harmonic sum
   S = +- 1/(a+1) +- 1/(a+2) +- 1/(a+3) +- ... +- 1/(a+n)
  where one puts any sign on any term, and there is at least one term. Let G be the LCM of the denominator(s). Then for any denominator D, G/D is an integer, and G*S = sum(+- G/D_i) is an integer. Let E be the highest power of two that divides G. Then there is only one multiple of E among the denominators. (If there were two, they would be consecutive multiples of E, and one would be divisible by 2*E.) Call that denominator F. So (+- G/F) is an odd integer, and for all other denominators D, (+- G/D) is an even integer. Therefore G*S is odd, therefore not zero, so S is not zero. (End)

			0, 1, 1/2, 1/6, 5/12, 13/60, 1/20, 27/140, 19/280, 451/2520, 199/2520, 4709/27720, ...

David Wilson, Posting to Sequence Fans Mailing List, Nov 14 2013.

David W. Wilson and Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A231693, A005132.

a231692_list = map numerator $ 0 : wilson 1 0 where
   wilson x y = y' : wilson (x + 1) y'
                where y' = y + (if y < 1 % x then 1 else -1) % x
-- Reinhard Zumkeller, Nov 16 2013

f:=proc(n) option remember;
if n=0 then 0 elif
f(n-1) >= 1/n then f(n-1)-1/n else f(n-1)+1/n; fi; end;

Numerator[FoldList[# + (-1)^Boole[#*#2 >= 1]/#2 &, Range[0, 35]]] (* Paolo Xausa, Aug 16 2025 *)

s=0;vector(30,n,numerator(s-=(-1)^(n*s<1)/n)) \\ - M. F. Hasler, Nov 15 2013

A232090 Minimal possible denominator for a sum of the form 1 +/- 1/2 +/- 1/3 +/- ... +/- 1/n.

Original entry on oeis.org

1, 2, 6, 12, 60, 20, 140, 280, 2520, 2520, 27720, 27720, 360360, 360360, 72072, 144144, 2450448, 272272, 5173168, 5173168, 739024, 739024, 16997552, 16997552, 424938800, 424938800, 11473347600, 11473347600, 332727080400, 332727080400

Offset: 1

M. F. Hasler, Nov 18 2013

nonn

Differs from A203811 from a(18)=272272 on, and from A002805 and A231693 from a(15)=72072 on.

Giovanni Resta, Table of n, a(n) for n = 1..37
F. T. Adams-Watters, Re: Reciprocal Recaman, SeqFan list, Nov 17 2013

Cf. A061195 (minimal possible positive numerator).

nMax = 19; d = {0}; Table[d = Flatten[{d + 1/n, d - 1/n}]; Min[Denominator[d]], {n, nMax}] (* T. D. Noe, Nov 19 2013 *)

for(n=0,19,m=(n+1)!;for(k=0,2^n-1,m=min(denominator(sum(j=2,n+1,(-1)^bittest(k,j-2)/j,1)),m));print1(m","))

Terms a(21)-a(30) from David W. Wilson, Nov 19 2013

A386883 Define a sequence of rationals by f(0) = 0, thereafter f(n) = f(n-1) - 1/n if that is >= 0, otherwise f(n) = f(n-1) + 1/n; a(n) corresponds to the number of addition steps minus the number of subtraction steps involved in calculating f(n).

Original entry on oeis.org

0, 1, 0, -1, 0, -1, -2, -1, -2, -1, -2, -1, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3

Offset: 0

Rémy Sigrist, Aug 06 2025

This sequence is unbounded below.
By contradiction:
- let M be the minimum value of the sequence and a(n) = M be the first occurrence of M in the sequence,
- as an addition step is always followed by a subtraction step, and the value M can only be followed by M+1, a(n+2*k) = M and a(n+2*k+1) = M+1 for any k >= 0,
- for any m >= n, f(m) = f(n) + Sum_{k = n+1..m} (-1)^(k-n-1) / k,
- as the alternating harmonic series converges to log(2), the sequence f will have a positive limit, say L > 0,
- hence for some m0 >= 0, f(m) > L/2 for any m >= m0,
- let k be such that n+2*k >= m0 and 1/(n+2*k+1) < L/2: f(n+2*k) > L/2, so f(n+2*k+1) = f(n+2*k) - 1/(n+2*k+1), and a(n+2*k+1) = a(n+2*k)-1 = M-1, a contradiction.

			Sequence begins:
  n   a(n)  f(n)-f(n-1)
  --  ----  -----------
   0     0  N/A
   1     1  +1
   2     0  -1/2
   3    -1  -1/3
   4     0  +1/4
   5    -1  -1/5
   6    -2  -1/6
   7    -1  +1/7
   8    -2  -1/8
   9    -1  +1/9
  10    -2  -1/10
  11    -1  +1/11
  12    -2  -1/12
  13    -3  -1/13
  14    -2  +1/14
  15    -3  -1/15

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A002162, A231692, A231693.

Module[{f = 0, a = 0}, Array[If[f >= 1/#, f -= 1/#; a--, f += 1/#; a++] &, 100]] (* Paolo Xausa, Aug 25 2025 *)

{ print1(0); f = 0; a = 0; for (n = 1, 65, if (f >= 1/n, f -= 1/n; a--, f += 1/n; a++); print1 (", "a);); }

a(n) = Sum_{k = 1..n} sign(A231692(n)/A231693(n) - A231692(n-1)/A231693(n-1)).

cp's OEIS Frontend

A231692 Define a sequence of rationals by f(0)=0, thereafter f(n)=f(n-1)-1/n if that is >= 0, otherwise f(n)=f(n-1)+1/n; a(n) = numerator of f(n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

A232090 Minimal possible denominator for a sum of the form 1 +/- 1/2 +/- 1/3 +/- ... +/- 1/n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A386883 Define a sequence of rationals by f(0) = 0, thereafter f(n) = f(n-1) - 1/n if that is >= 0, otherwise f(n) = f(n-1) + 1/n; a(n) corresponds to the number of addition steps minus the number of subtraction steps involved in calculating f(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula